# Given the function f(x) and the identity 3f(x)-f(-x)=x+2, determine the area bounded by f(x), x and y axis and the line x=1?

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Given the identity 3f(x)-f(-x)=x+2, that is verified for any real value of x, then we can replace x by -x and we'll get an identity, also.

3f(-x) - f(x) = -x + 2

We'll create a system formed from the identities above:

-f(-x) + 3f(x) = x + 2 (1)

3f(-x) - f(x) = -x + 2 (2)

We'll eliminate f(-x) by multiplying (1) by 3 and adding the expression resulted to (2):

-3f(-x) + 9f(x) + 3f(-x) - f(x) = 3x + 6 - x + 2

We'll combine like terms:

8f(x) = 2x + 8

f(x) = x/4 + 1

Since we know the expression of the function f(x), we can evaluate the area under the curve f(x) and bounded by x and y axis and the line x = 1.

We'll calculate the definite integral of f(x), whose limits of integration are x = 0 and x = 1.

Int f(x)dx = Int (x/4 + 1)dx

Int (x/4 + 1)dx = (1/4)*Int xdx + Int dx

Int f(x)dx = x^2/8 + x

We'll apply Leibniz Newton:

Int f(x)dx = F(1) - F(0)

F(1) - F(0) = 1/8 + 1 - 0/8 - 0

F(1) - F(0) = 9/8

**The area bounded by the curve f(x) = x/4 + 1, the x lines and the limits x = 0 to x =1, is A = Int f(x)dx = 9/8 square units.**